Abstract

Let and let be an algebraically closed field with characteristic 0 or greater than . We show that if and satisfy , then are simultaneously triangularizable. Let be a reduced ring such that is not a zero divisor and let be a generic matrix over ; we show that is the sole solution of . Let be a commutative ring with unity; let be similar to such that, for every is not a zero divisor. If is a nilpotent solution of where , then .

1. Introduction

Let be an integer at least . In the first part, we consider matrices with entries in , a field such that its characteristic is or greater than , and denotes its algebraic closure. In the second part, the entries of the matrices are elements of , a commutative ring with unity ; then, is the ring of endomorphisms of the free -module .

Let and let be a polynomial with coefficients in the complex field such that . In [1, 2], the matrix is given and the authors study the matrix equations in the unknown :

In the present paper, we extend some results obtained in [1, 2] when we replace with or .

Two matrices , are said to be simultaneously triangularizable (abbreviated to ) over if there exists such that and are upper triangular matrices.

The following result is a slight improvement of [1, Theorem 1] or of [3, Theorem 11′].

Theorem 1. One assumes that or is and that is a polynomial with coefficients in :
Then, , are over .

Remark 2. Note that the previous result fails when .

We now consider (1) and (2).

Let be a commutative ring with unity and let be a matrix where the are commuting indeterminates. Let be the ring of the polynomials in the indeterminates and with coefficients in . Thus, the algebra generated by is included in . In particular, there are no polynomial relations, with coefficients in , linking the . We say that is a generic matrix over . When is reduced (for every , implies ), we obtain a precise result.

Proposition 3. Let , and let be a reduced ring such that is not a zero divisor. Let be a generic matrix. Then, is the sole solution of (1).

Otherwise, we only obtain a partial result.

Proposition 4. Let be a commutative ring with unity, , and let be a polynomial with coefficients in such that . Let be a nilpotent solution of (2). Then, all elements of the two-sided ideal, in , generated by are nilpotent.

Yet, if is diagonalizable and its spectrum is “good,” then we obtain a complete solution.

Theorem 5. Let be similar over to with such that, for every , is not a zero divisor. Let be a nilpotent solution of (2). (i)Then, there is such that where, for every , and .(ii)If moreover is a unit, then .

2. Equation (1) Over a Field

In this section, is a field with characteristic not . Let denote the nilpotent Jordan-block of dimension . If is a square matrix, then denotes its characteristic polynomial and denotes its commutant.

We prove Theorem 1.

Proof. Let be the vector space spanned by . One checks easily by induction:
By Cayley-Hamilton’s Theorem (that is valid on a commutative ring with unity), belongs to , and is Lie’s algebra. The derived series of is
Thus, is solvable. According to Lie’s Theorem (that is valid when or is , cf. [4, page 38]), is triangularizable; that is, , are .

Remark 6. The hypothesis about is necessary; indeed, if and , then
satisfy ; yet, is not nilpotent.

Example 7. Consider the following equation:
It is equivalent to
According to Theorem 1, are over and is nilpotent.

Proposition 8. Let and let  be such that its eigenvalues in have multiplicity at most . Then, the solutions of (1) commute with .

Proof. According to [2, Proposition 2.9], if is a solution of (1), then each generalized eigenspace of is -invariant. Thus, we may assume that has dimension at most . Therefore, and we are done.

We consider the relation:

Proposition 9. One assumes that or is . Let satisfying (10) such that . Then , is nilpotent and there is and a nilpotent matrix such that ; moreover, . Conversely, there exist such solutions.

Proof. If , then, according to Proposition 8, that is contradictory. Thus, and has a triple eigenvalue . An instance of such a solution is

Let . According to [2, Proposition 2.9], if is a solution of (1), then each generalized eigenspace of is -invariant. Thus, we may assume that is nilpotent.

We consider the algebraic variety:

Let denote the dimension of the maximal component of . Recall that the algebraic variety of nilpotent matrices in has dimension and is irreducible (cf. [5, Section: The nilpotent cone]). Note that the algebraic variety

has dimension and is irreducible when (cf. [6]).

Proposition 10. Let . Then, .

Proof. A generic nilpotent matrix is similar to . Put and consider the equation . According to [2, Remark 3.4], is strictly upper triangular and we can express the entries of as functions of the free variables (the only condition occurs when and is ). Then, the algebraic variety has dimension . We deduce that .

Remark 11. We show that (cf. Proposition 9).

Case 1. . Then, ; that is, or is similar to ; we obtain a variety of dimension .

Case 2. is similar to ; that is, goes through a variety of dimension  . From the Gröbner basis method (implemented in Maple), we deduce that the Hilbert dimension of is . Thus, the dimension of the component of associated to this choice of is .

Computational evidence leads us to conjecture that , an expression that does not depend on .

We consider the equation:

A family of such equations is studied in [7]. As in [1, 2], an essential argument used by the authors is that the sequence of iterated kernels is ultimately stationary. Moreover, the properties about the generalized eigenspaces are similar; and have same spectrum over ; moreover, for every , the generalized eigenspaces and are equal. Thus, to study the solutions of (14) can be reduced to study the restrictions of to a generalized eigenspace . Moreover, if is a solution of (14), then, for every , is also a solution of (14). Finally, it suffices to solve (14) when , are assumed to be nilpotent matrices.

Let be a solution of (14). We may wonder whether , are . The answer is no, as we can see with the following solution of (14) when and or :

Clearly , are nilpotent and is invertible. We say that a pair has property (cf. [8]) if there are orderings of the eigenvalues of , such that, for every , the eigenvalues of are ; if , are , then they have property . In our instance, has not property because, if , then is not nilpotent.

Let be integers .

Proposition 12. Let be a field such that or is . If has distinct eigenvalues in , then is the sole solution of (1).

Proof. Note that satisfies the property:
Indeed, we may assume that is a diagonal matrix over . Since commutes with , is also diagonal and clearly . According to Theorem 1, is nilpotent and too; assume that , the nilindex of , is greater than . According to (5), ; by the property and , we deduce , that is contradictory and therefore . Thus, and ; by Property , we conclude that .

Remark 13. The previous result is shown, when is a field of characteristic , in [2, Proposition 2.5].

3. Equations (1) and (2) Over a Ring

(i)Let be commutative rings with unity. Their ring subdirect product is defined if there is an injective ring homomorphism such that, for every , the projection of on is onto.(ii)A commutative ring with unity is reduced if, for every , implies . That is equivalent to say that is isomorphic to a subring of a direct product of fields or isomorphic to a subdirect product of domains (cf. [9, Theorem 11.6.7]). For instance, is a reduced ring with . More generally, is or a product of distinct primes. Note that is reduced with and yet, is a zero divisor.

We show Proposition 3.

Proof. Note that, if is reduced, then also is. Since is a subring of a direct product of algebraic closed fields , we may assume where, for every , is a field such that or is . Let and be the associated decompositions. Thus, for any , , where the th component of is generic; then, for every , the discriminant of is not and the matrix has distinct eigenvalues. According to Proposition 12, for every , and consequently .

Proposition 14. Let be a commutative ring with unity such that is not a zero divisor and let . If is a solution of (1), then is a nilpotent matrix.

Proof. Note that and commute and that the Cayley-Hamilton theorem is true over . According to the proof of Jacobson lemma (cf. introduction of [10] and also [11] where an improvement of this result is stated within the framework of the algebraic operators on a complex Banach space), that implies and we are done.

If is generic over , then we have a more precise result for small .

Proposition 15. Let , . Let be a commutative ring with unity such that, if , , or , then , or is not a zero divisor. Let be a generic matrix. If is a solution of (1) then

Proof. The parameters are the . We have a system of equations in the unknowns . Using Gröbner basis theory in any specified characteristic, we obtain the required result.

When and , the calculations have great complexity; thus, we carry out specializations of the in the ring . Then, we randomly choose the matrix (in order to simulate the generic nature of the matrix) and we formally solve (1) in the unknowns . Numerical experiments, again using Gröbner basis theory in characteristic great enough, lead to the following result: and for every , ; for instance, if , , then the supplementary condition is “ is not a zero divisor.” Therefore, we conjecture the following.

Conjecture. Let , be a commutative ring with unity satisfying a condition in the form: “the integer is not a zero divisor.” Let be a generic matrix. If is a solution of (1), then and for every , .

Remark 16. (i) The instance  , shows that if is a nilpotent matrix, then we have not necessarily .
(ii) In the previous conjecture, note that the exponent is very special; indeed, if , then we have
and we cannot do better (cf. [12]).

Let be a commutative ring with unity and . We look for the nilpotent solutions of (2), where and is a polynomial in , with coefficients in , such that . Then, according to (5), for every ,

Let denote the valuation of the polynomial , with the following convention: . In the sequel, is a nilpotent solution of (2).

Lemma 17. Let be a polynomial in . Then, for every , , where for every , is a polynomial in such that .

Proof. For every , with . Then with . In the same way, and with , and so on.

Lemma 18. Let and be polynomials in with, for every , . Then
where for every , is a polynomial in with .

Proof. We use a reasoning by recurrence. Let
where for every , is a polynomial in such that . Using Lemma 17, , where .

Lemma 19. Let be polynomials in . Then, for every ,
where for every , is a polynomial in such that .

Proof. Let . Then , where for every , are monomials in the form . Assume, for instance, that . By Lemma 17,
and finally , are in the form , , where , are polynomials in . Now
where for every , . Using Lemma 18, , where for every , .

We deduce Proposition 4.

Proof. Use Lemmas 18 and 19 and the fact that is a nilpotent matrix.

Remark 20. (i) When is an arbitrary algebraically closed field, the previous result is equivalent to say that and are over (cf. [13] and compare with Theorem 1).
(ii) When is a ring, McCoy, in [14], gave an equivalent condition that, unfortunately, seems almost useless. In fact, if and , are triangularizable over , then they have not necessarily a common eigenvector; the following example, for , is due to J. Starr and uses the ring of dual numbers:
Moreover, let ; then, unlike the case where the entries of the matrices are in a field, the equation admits a solution that does not commute with ; yet .

Note that we can reduce (theoretically) the resolution of (10) to the case . Indeed, let be satisfying (10). According to (5), . If is a unit, then put , ; therefore, .

We have a more precise result when is diagonalizable and its spectrum is “good.”

Lemma 21. Let be similar to and such that . (i)Assume that, for every , is not a zero divisor in . Then and are simultaneously diagonalizable.(ii)Assume that, for every , is a unit. Then is a polynomial in of degree at most and with coefficients in .

Proof. We may assume that . (i)If , then with ; if , then .(ii)According to (i), we may assume that . We must solve the linear system, in the unknowns :
Since the determinant of the associated Vandermonde matrix is a unit, we are done.

Proposition 22. Let be similar to such that, for every , is not a zero divisor and . If and commute, then .

Proof. We may assume that and put , . We obtain ; therefore, if , then .

The proof of the following is interesting in itself; yet, Theorem 5 is a stronger result.

Proposition 23. Let be similar in to such that, for every , is not a zero divisor. If is not a zero divisor and is a solution of (1), then there are   and such that